Answer
Please see the work below.
Work Step by Step
We know that the range of the first projectile is given as:
$R_1=\frac{v_{\circ}^2 sin(2\theta)}{g}$
$R_1=\frac{v_{\circ}^2 sin(2(45+\alpha))}{g}=\frac{v_{\circ}^2 sin(90+2\alpha)}{g}$
We can find the range of the second projectile as:
$R_2=\frac{v_{\circ}^2 sin(2\theta)}{g}$
$R_2=\frac{v_{\circ}^2 sin(2(45-\alpha))}{g}=\frac{v_{\circ}^2 sin(90-2\alpha)}{g}$
As we know that $sin(90+\theta)=sin(90-\theta)$, the range of the projectile is the same.
